Atwoods Analysis

[pequaide]

I calculate 1.3 metric tons dropped one meter.

I drop about a tenth of that just a few centimeters. So I know about how much 1.3 metric tons one meter would do.

If torque is newton meters it is way more out than in. But this is Wubbly's site and it is time to move on from 'can't be done' territory.

[Wubbly]

pequaide, you still seem to be confusing torque with moment of inertia. You had them confused in Tarsier's mrr thread, and you still haven't quite got it.

Confusing torque with moment of inertia is like confusing force with mass.

In the Atwoods, one set of masses (the driver masses) are creating the input torque (and are not being rotated), and the wheel or pulley mass is the mass being rotated.

In the mass/rod experiment, there is one mass that is creating the input torque and is also the same mass that is being rotated.

The purpose of the mass/rod experiment was to show you that I=mrr was real using a very simple example. path_finders picture showed a clever experiment that allowed you to change the position of the mass (and observe changes to moment of inertia) without changing the input torque of the pendulum.

Look at the attached pictures of various Atwoods.

Use a zero pully mass. For your driver masses, use a constant mass difference of 1. IMO, This will create a constant input torque into your system. Increase your driver masses by doubling, tripling, and quadrupling the masses.
What happens to your acceleration. It becomes 1/2, then 1/3, then 1/4.

This means that your mass is acting like regular mass. It is not following the mrr relationship.
Why not?
It's because IN AN ATWOODS, THE DRIVER MASSES ARE NOT BEING ROTATED.

[Wubbly]

After thinking about it some more, that last conclusion in that last post was incorrect. Using those numbers, you can't determine if the masses are following an mrr relationship. Using those numbers, you get the following relationships:

T = I x alpha

T = torque
I = MOI (moment of inertia)
alpha = angular acceleration

Code: Select all

1T = 1I x   1 alpha
1T = 2I x 1/2 alpha
1T = 3I x 1/3 alpha
1T = 4I x 1/4 alpha

You can, however, determine that the mrr relationship holds true in an Atwoods using a 2 kg mass on one side, and a 1 kg mass on the other side. Vary the radius from 1m to 2m to 3m to 4m. Now your input torque is varying between the experiments as the radius changes, but the total mass (3 kg) stays the same. Also note that as you increase your radius, the linear acceleration will stay the same, but the angular acceleration will decrease. Now you have the following relationships:

Code: Select all

1T =  1I x   1 alpha
2T =  4I x 1/2 alpha
3T =  9I x 1/3 alpha
4T = 16I x 1/4 alpha

If you apply twice the torque and get half the acceleration, your mass is acting like it's 4 times heavier to satisfy the equation. Yeah, now it makes sense. The radius in the mrr is not the distance from the center of the pully to the mass, it is the radius of the pully itself.

[Wubbly]

How to make a 50 kg flywheel feel like it is 11,250 kilograms:

Lets suppose you have a 50 kg flywheel with a radius of 0.3 meters and most of its mass concentrated at the rim.

It would have a Moment of inertia "I" of: I=mr² = 50 kg * (0.3 m)² = 4.5 kg m²

The linear acceleration of the driver mass m1, hanging off of a flywheel at a radius r1 is: a = g * m1 / (m1+ I/r1²)

The I/r1² term in the denominator is what the mass of the flywheel contributes to the equation.
If r1 = 0.3m, then to the driver mass, the flywheel feels like it's 50 kg.
If r1 is half of 0.3m, then to the driver mass, the flywheel feels four times heavier (200 kg).
As r1 gets samller, the flywheel feels exponentially heavier to the driver mass.
So at a driver radius of 0.02 m, to the driver mass the flywheel feels like it is 11,250 kg.

If your driver mass was 3 kg, It could exert a force of 3 kg * 9.81 m/s² = 29.43 N.

In a linear system, If your force is 29.43 N, and is accelerating a mass of 11,250 kg, your acceleration would be:
a = F/m = 29.43N / 11,250 kg = 0.002616 m/s²

If you plug the mass, radius, and moment of inertia numbers into the acceleration equation for our rotational system a = g * m1 / (m1+ I/r1²) you get an acceleration of 0.002615 m/s².

You get almost the same acceleration.

Conclusion: To the 3 kg driver mass at a radius of 0.02 meters, the 50 kg flywheel feels like it is 11,250 kg.

The I/r1² term in the denominator contributes significantly to how the system accelerates.

[Wubbly]

.
In the previous example with the 50 kg flywheel, what would happen if, in addition to the 3 kg driver mass, we hung 21 kg off of each side at the same radius as the driver mass (0.02m)?
Now the mass we are accelerating would be the 3 kg driver mass, plus 21 kg + 21 kg extra hanging mass, plus the 50 kg flywheel (that is acting like it is 11,250 kg).
So the total mass we are accelerating would be 3 + 21 + 21 + 11,250 = 11,295 kg.
Our force accelerating the system is 9.81 m/s² * 3 kg = 29.43 N
So the linear acceleration of the driver mass would be a=F/m = 29.43 N / 11,295 kg = 0.002606 m/s².

So even though we added 42 kg to our system, the linear acceleration of the driver mass only changed by 0.000009 m/s² from the acceleration in the previous post.

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Now lets throw away the two 21 kg extra hanging masses on each side and replace it with 1.4 kg extra hanging mass on each side, but at a radius of the flywheel (0.3m).
A 1.4 kg mass is one fifteenth the mass of a 21 kg mass, but it is hanging at a radius fifteen times greater than the driver mass (0.02 * 15 = 0.3)
To the driver mass, each 1.4 kg mass at fifteen times the radius would feel like it is 1.4 kg x 15² = 315 kg.
So the total mass we are accelerating would be 3 + 315 + 315 + 11,250 = 11,883 kg.
Our force accelerating the system is still the same 9.81 m/s² * 3 kg = 29.43 N
So the new linear acceleration of the driver mass would be a=F/m = 29.43 N / 11,883 kg = 0.002477 m/s².

From the previous experiment (with the 21 kg + 21 kg hanging at a radius of 0.02 m), to this last experiment (with 1.4 kg + 1.4 kg hanging at a radius of 0.3 m), the linear acceleration of the driver mass only changed by 0.000129 m/s².

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The massive flywheel is the elephant in the experiment that dominates the acceleration, and the stuff you are hanging off of it is down in the weeds.

The obvious conclusion is that if one were making comparisons of the relative accelerations between hanging masses in an Atwoods, one could not possibly find a worse experimental setup than what is described in these last two posts.
.

[Tarsier79]

Wubbly, Im not sure if you should count the driver mass in your calculations. Gravity pulls the driver down, which in turn adds rotational force to the wheel, as would someone pulling down on the string supplying the same force as the weight.

When there is a driver weight at a different radius than the inertial weights, I suspect you have to find the MOI of the whole system, then there will be a mathematical relationship between the ratios of the driver and inertial mass that you have to balance when you calculate the new apparent MOI.

I have not had a chance to test, but this seems logical to me.

[Wubbly]

.
I think you would have to count the driver mass in the calculations because it shows up in the derivation of the acceleration equation.
Attached is a derivation of the linear acceleration equation with different radii between the driver mass and the other hanging masses. I thought that the "I" in the equation only came from the moment of inertia of the pully. You don't have to calculate the MOI of the hanging masses but a term similar to mr² appears when you work out the algebra.

Here's how the acceleration equation is derived:
(1) sum the forces on mass 1. => solve for T1
(2) sum the forces on mass 2. => solve for T2
(3) sum the forces on mass 3. => solve for T3
(4) Radius 2 is some constant times radius 1.
(5) Acceleration 2 of the outer masses is a constant times the acceleration of the driver mass. Plug this into (2) and (3) to eliminate a2.
(6) Angular acceleration = linear acceleration a1/ radius1
(7) The sum of the torques on the flywheel is equal to the moment of inertia of the flywheel times the angular acceleration.
Plug (6) into (7) to eliminate alpha.
Plug (4) into (7) to eliminate r2.
Plug (1) (2) and (3) into (7) and solve for linear acceleration a1.

note:
The driver mass m1 does not disappear from the equation.
When working out the algebra, a distance squared term magically appears in front of m2 and m3. The "C" is how much larger r2 is than r1.

Using tension analysis and Newton's laws of motion one can derive the acceleration equation for various Atwoods configurations.
.

[Wubbly]

Attached is a spreadsheet for analyzing various Atwoods machines with from 1 to 4 hanging masses. You can change masses, heights, etc. and see how it affects the acceleration. There are NO macros in this spreadsheet, so if it warns you about enabling macros, then that is NOT the spreadsheet I uploaded. I saved it in the old 2003 format to make it more compatable on older computer systems.

Nothing can be found in an Atwoods machine that produces any sort of energy gain. You are simply dropping masses and exchanging potential energy for kinetic energy.